Answer to Question #126361 in Physics for Abdul Rashidul

Question

An ice cube of mass, 75 g at -10.0 °C is placed in 0.5 kg of water at 50.0 °C in an insulating
container. Determine
a. whether the ice will melt completely or not
b. the final temperature of the system
[cice= 2.1 kJ/kg.K, Lf of ice is 333 kJ/kg, cwater = 4.2 kJ/kg.K]

Answer 1

The equation of termal balance of the system will be the following:

$$Q_{melt} + Q_{ice} +Q_{cold}= Q_{water}$$

where $Q_{melt}$ is the energy required to melt the ice, $Q_{ice}$ is the energy required to heat the ice to the melting point (273 K), $Q_{cold}$ is the energy required to heat the melted ice to the final temperature and $Q_{hot}$ is the energy spent by the hot water.

Let's define each term.

$$Q_{melt} = L_fm_{ice} = 24.975\times10^3J$$

where $L_f = 333\times 10^3J/kg$ is the latent heat of fusion of ice, and $m_{ice} = 0.075kg$ is the mass of ice.

$$Q_{ice} = c_{ice}m_{ice}(273K - T_{ice}) = 1.575\times 10^3J$$

where $c_{ice} = 2.1\times 10^3J/kg.K$ is the specific heat capacity of ice, $T$ is the final temperature of the system and $T_{ice}$ is the initial temperature of the ice.

$$Q_{cold} = c_{water}m_{ice}(T - 273K)$$

where $c_{water} = 4.2\times 10^3J/kg.K$ is the specific heat capacity of water.

Similarly, the energy spent by the water will be:

$$Q_{water} = c_{water}m_{water}(T_{water}-T)$$

Now let's substitute all these expression into the initial equation:

$$24.975\times10^3J + 1.575\times 10^3J + c_{water}m_{ice}(T - 273K) = c_{water}m_{water}(T_{water}-T)\\ 26.55\times10^3 + c_{water}m_{ice}T - c_{water}m_{ice}273K = c_{water}m_{water}T_{water}- c_{water}m_{water}T$$

Dividing both sides by the c_{water}, obtain:

$$26.55\times10^3/c_{water} + m_{ice}T - m_{ice}273K = m_{water}T_{water}- m_{water}T\\ T(m_{ice} + m_{water}) = m_{water}T_{water} +m_{ice}273K - 26.55\times10^3/c_{water}\\ T = \dfrac{m_{water}T_{water} +m_{ice}273K - 26.55\times10^3/c_{water}}{m_{ice} + m_{water}}$$

Substituting the numerical values, obtain:

$$T = \dfrac{0.5\cdot323 +0.075\cdot273 - \frac{26.55\times10^3}{4.2\times10^3}}{0.075 + 0.5} \approx 305.5K = 32.5\degree C$$

Answer. (a) the ice will melt completely, (b) the final temperature is 32.5 C.